Answer :
Answer:
(-2, 6)
Step-by-step explanation:
we have
[tex]y=x^{2} +3x+8[/tex] ----> equation A
[tex]y=2x+10[/tex] ----> equation B
equate equation A and equation B
[tex]x^{2} +3x+8=2x+10[/tex]
[tex]x^{2} +3x+8-2x-10=0[/tex]
[tex]x^{2} +x-2=0[/tex]
The formula to solve a quadratic equation of the form
[tex]ax^{2} +bx+c=0[/tex]
is equal to
[tex]x=\frac{-b\pm\sqrt{b^{2}-4ac}} {2a}[/tex]
in this problem we have
[tex]x^{2} +x-2=0[/tex]
so
[tex]a=1\\b=1\\c=-2[/tex]
substitute in the formula
[tex]x=\frac{-1\pm\sqrt{1^{2}-4(1)(-2)}} {2(1)}[/tex]
[tex]x=\frac{-1\pm\sqrt{9}} {2}[/tex]
[tex]x=\frac{-1\pm3} {2}[/tex]
[tex]x=\frac{-1+3} {2}=1[/tex]
[tex]x=\frac{-1-3} {2}=-2[/tex]
Find the values of y
For x=1
[tex]y=2(1)+10=12[/tex]
For x=-2
[tex]y=2(-2)+10=6[/tex]
therefore
The solutions are (1,12) and (-2,6)